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https://web2.0calc.com/questions/rolls_2

Sorry....see my corrected answer

I've seen this problum so i'm assuming that the ratio is 3:2, That means if everyone has an equal chance to be president, the ratio of sophomore swimmers to freshman swimmers is 3:2. So 3/(3+2) is 3/5 of the team are sophomores.

The probablility for the first card to be a king is 1/13

the probability for the second card to be hearts is 3/4 likely to be 1/4 or and 1/4 likely to be 3/13

The probablity for the third to be red is 3/4 likely 1/2 or 1/4 likely 6/13

(1/13*3/4*1/4 +1/13*1/4*3/13) *3/4*1/2+

(1/13*3/4*1/4 +1/13*1/4*3/13) *1/4*6/13

=

2601/281216​

...really?

First of all, at least try the problem before copy and pasting it onto this site

Second of all, you didn't even copy it RIGHT. I would expect you put a little more effort so that people can READ your problem, and answer it occordingly. 

You can post it again and I will answer it. Post it with the corrections.

We can imagine a coordinate grid The right-most point he can travel to is (2,0), left-most is (-2,0), top and down is (0,2) and (0,-2) respectively.

Now, in between those points, the furthest the fly can go is (1,1),(1,-1),(-1,1), and (-1,-1). Therefore, the total number of possible points Franklin can end up at is \(\fbox{12}\)

The fisrt answer was right... but cphill's answer was incorrect

True....but i'm assuming that x and y are different integers.....

However....if  let them  be the same, then your answer is  correct

2. 

Let ABCDEFGH be right rectangular prism. The total surface area of the prism 30. Also, the sum of all the edges of the prism is 44. Find the length of the diagonal joining one corner of the prism to the opposite corner.

h = height

w = width

l = length

4 ( h + w + l)  =  44

h + w +  l =  11                   (1)

2 (lw + lh + hw) = 30          (2)

Square both sides of (1)

h^2 + 2 h l + 2 h w + l^2 + 2 l w + w^2  = 121       rearrange as

h^2 + l^2 + w^2 +  2 (lw + lh + hw)  = 121  (3)

Sub (2)  into (3)

h^2 + l^2 + w^2  +  30 =  121

h^2 + l^2 + w^2  = 91

sqrt  [ h^2 + l^2 + w^2 ] = sqrt (91)  ≈  9.54  =   length of the diagonal joining one corner of the prism to the opposite corner

CORRECTED

Set the equations equal

3  = 4x^2 + x  - 1         simplify

4x^2 + x - 4  = 0        divide through by 4

x^2  - x/4  - 1  = 0

x^2 + x/4  =   1                complete the square on x

x^2  + x/4  + 1/64  =  1 + 1/64

(x + 1/8)^2  = 65/64            take both roots

x + 1/8  =  sqrt(65) /  8                       x +  1/8  = -sqrt (65) / 8

x =   [ sqrt (65) - 1 ]  / 8                  x =  [-sqrt (65) - 1 ]  / 8

A =  ( [-sqrt (65) - 1 ] / 8 , 3 )

B = ([ sqrt (65) - 1 ] / 8 , 3 )

The distance between  these points =  2sqrt (65) / 8  =   sqrt (65)  /  4

m - n =  65  - 4  =   61

\(400000a+40000b+4000c+400d+40e+4f=300000d+30000e+3000f+300a+30b+3c\)

\(3997(\overline{abc})=2996(\overline{def})\)

\(\overline{abcdef} = 29963997\)

Alrighty..

1. For this one, the diagonal of the base ABCD is \(\sqrt{12^2+12^2}\) which is \(12\sqrt2\). Dividing that by 2 gives \(6\sqrt2\).

Now, to find the height, we use the pythagorean theroem.

\(10^2={6\sqrt2}^2+b^2\)

\(100 = 72+b^2\)

\(b^2 = 28\)

\(b = 2\sqrt7\)

by the way, b is the height.

now, we get the area of the pyramid is \(\frac{12\cdot12\cdot2\sqrt7}{3}\)

which gives us \(96\sqrt7\)

\(2i\)

:)

30 + 30 is 60   

_.

3a + 2b =   2       (1)     

a + 2b + 8a + mb  = k   →   9a + (2 + m)b   =  k    (2)

The system has infinite solutios when the equations are exactly the same

Multiply (1)  by 3

9a + 6b  =  6       (3)

Equating terms in (2) and (3)

(2 + m)  = 6           k =  6    

m  =  4

https://web2.0calc.com/questions/simplify_39168

1/x  + 1/y =  1/15

( x + y)  / (xy)  = 1/15

(xy)    / ( x + y)  =  15

xy =  15 ( x + y)

xy  =15x + 15y

xy - 15y =  15x

y ( x - 15)  =  15x

y =   15x / ( x - 15)

x          y

16      240

18       90

20       60

24       40

30       30

40       24

Smallest  x + y =  40 + 24  =  64

thank you so much!! 

Well there's two cubes, so the smallest possible configuration of them would be a rectangular prism...

So we have a rectangular prism with height 1, width 1, and length 2. 

Surface area would be 10.

Correct me if im wrong.

Use 3d pythagorean theorem,

3^2+5^2+11^2=x^2.

9+25+121=x^2

155=x^2

x=sqrt 155

So the distance between the two opposite corners is sqrt 155

zw - 3w - 2iw + 4iz = -8 + 12i

Where did you get this equation?

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The pyramid  EFGH  will  have the same base and the  sme height as pyramid DEFG....so....its volume also   =  18

-6x + 3y  = -18

3y = 6x -18      divide through by 3

y = 2x - 6

The slope of the line we're looking for = 2

x intercept  of  14x +7y - 14  =0

Let y = 0

14x - 14 =  0

14x  =14    divide through by 14

x =1

So.....we have the point (1,0)

Equation of line

y = 2 (x -1)

y = 2x  - 2

2x - y  - 2   =  0

\(zw-3w-2iw=4iz=-8+12i\)

\(4iz=-8+12i\\ z=\frac{-8+12i}{4i}\\ \color{blue}z=3+2i\)

\(zw-3w-2iw=4iz\\ (3+2i)w-3w-2iw=4i(3+2i)\\ w(3+2i-3-2i)=-8+12i\\ w= \frac{-8+12i}{0}\\ \color{blue}w=indefinite\)

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